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5
votes
Accepted
Borel $\sigma$-algebra in $\beta \mathbb N \times \beta \mathbb N$
The answer is no.
Jiří Nedoma proved that if $(X,\Sigma)$ is a measurable space $|X| > 2^{\aleph_0}$, then the diagonal is not a measurable subset of $(X\times X, \Sigma \otimes \Sigma)$. (The articl …
3
votes
Accepted
The space of Borel function modulo comeager sets is Dedekind complete
Fremlin's measure theory textbook is a good reference for these things. I am splitting things up into the Boolean algebra part and the real-valued functions part.
Complete Boolean algebras:
The way …